Integrand size = 119, antiderivative size = 27 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\left (3-\frac {5}{3} e^{\frac {e^x}{x \left (x+x \log \left (\frac {1}{x}\right )\right )}}\right )^2 \] Output:
(3-5/3*exp(exp(x)/(x+x*ln(1/x))/x))^2
Time = 3.65 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {10}{9} \left (-9 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {5}{2} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \] Input:
Integrate[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 9 0*x)*Log[x^(-1)]) + E^((2*E^x)/(x^2 + x^2*Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x*(-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Log[ x^(-1)]^2),x]
Output:
(10*(-9*E^(E^x/(x^2*(1 + Log[x^(-1)]))) + (5*E^((2*E^x)/(x^2*(1 + Log[x^(- 1)]))))/2))/9
Time = 3.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {7239, 27, 7259, 17}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (50 x-50)+e^x (50 x-100) \log \left (\frac {1}{x}\right )\right )}{9 x^3+9 x^3 \log ^2\left (\frac {1}{x}\right )+18 x^3 \log \left (\frac {1}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {10 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}+x} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right ) \left (-x-(x-2) \log \left (\frac {1}{x}\right )+1\right )}{9 x^3 \left (\log \left (\frac {1}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {10}{9} \int \frac {e^{x+\frac {e^x}{\left (\log \left (\frac {1}{x}\right )+1\right ) x^2}} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right ) \left (-x+(2-x) \log \left (\frac {1}{x}\right )+1\right )}{x^3 \left (\log \left (\frac {1}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 7259 |
\(\displaystyle \frac {2}{9} \int \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )d\left (-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{9} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )^2\) |
Input:
Int[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 90*x)*L og[x^(-1)]) + E^((2*E^x)/(x^2 + x^2*Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x* (-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Log[x^(-1) ]^2),x]
Output:
(9 - 5*E^(E^x/(x^2*(1 + Log[x^(-1)]))))^2/9
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])]}, Simp[c Subst[Int[(a + b*x^p)^m, x] , x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]
Time = 5.68 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
risch | \(\frac {25 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}}{9}-10 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}\) | \(34\) |
Input:
int((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*ln(1/x)+ x^2))^2+((-90*x+180)*exp(x)*ln(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2*ln( 1/x)+x^2)))/(9*x^3*ln(1/x)^2+18*x^3*ln(1/x)+9*x^3),x,method=_RETURNVERBOSE )
Output:
25/9*exp(-2*exp(x)/x^2/(ln(x)-1))-10*exp(-exp(x)/x^2/(ln(x)-1))
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25}{9} \, e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 10 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} \] Input:
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m="fricas")
Output:
25/9*e^(2*e^x/(x^2*log(1/x) + x^2)) - 10*e^(e^x/(x^2*log(1/x) + x^2))
Time = 0.39 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25 e^{\frac {2 e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}}}{9} - 10 e^{\frac {e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}} \] Input:
integrate((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x**2*l n(1/x)+x**2))**2+((-90*x+180)*exp(x)*ln(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x**2*ln(1/x)+x**2)))/(9*x**3*ln(1/x)**2+18*x**3*ln(1/x)+9*x**3),x)
Output:
25*exp(2*exp(x)/(x**2*log(1/x) + x**2))/9 - 10*exp(exp(x)/(x**2*log(1/x) + x**2))
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-\frac {5}{9} \, {\left (18 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} - 5\right )} e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \] Input:
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m="maxima")
Output:
-5/9*(18*e^(e^x/(x^2*log(x) - x^2)) - 5)*e^(-2*e^x/(x^2*log(x) - x^2))
Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-10 \, e^{\left (-\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} + \frac {25}{9} \, e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \] Input:
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m="giac")
Output:
-10*e^(-e^x/(x^2*log(x) - x^2)) + 25/9*e^(-2*e^x/(x^2*log(x) - x^2))
Time = 7.62 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}\,\left (5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}-18\right )}{9} \] Input:
int((exp((2*exp(x))/(x^2*log(1/x) + x^2))*(exp(x)*(50*x - 50) + log(1/x)*e xp(x)*(50*x - 100)) - exp(exp(x)/(x^2*log(1/x) + x^2))*(exp(x)*(90*x - 90) + log(1/x)*exp(x)*(90*x - 180)))/(18*x^3*log(1/x) + 9*x^3 + 9*x^3*log(1/x )^2),x)
Output:
(5*exp(exp(x)/(x^2*log(1/x) + x^2))*(5*exp(exp(x)/(x^2*log(1/x) + x^2)) - 18))/9
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {-10 e^{\frac {e^{x}}{\mathrm {log}\left (x \right ) x^{2}-x^{2}}}+\frac {25}{9}}{e^{\frac {2 e^{x}}{\mathrm {log}\left (x \right ) x^{2}-x^{2}}}} \] Input:
int((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*log(1/x )+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2* log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x)
Output:
(5*( - 18*e**(e**x/(log(x)*x**2 - x**2)) + 5))/(9*e**((2*e**x)/(log(x)*x** 2 - x**2)))